3 Marks Question

Question 13 Marks

Which of the following relationships between the acceleration $a$ and the displacement $x$ of a particle involve simple harmonic motion?
(a) $a=0.7 x$
(b) $a=-200 x^2$
(c) $a=-10 x$
(d) $a=100 x^3$

Answer

If the acceleration $a$ fulfills the following conditions then we will say that the particle S.H.M.
(i) $a \propto x$
(ii) $a=-\omega^2 x$$\ldots$(1)
Here acceleration $a$ and displacement $x$ which is taken from the mean position.
(a) $a=0.7 x$ does not satisfy equation (1), hence the particle does not represent S.H.M.
(b) $a=-200 x^2$ does not satisfy equation (1), hence this does not represent S.H.M.
(c) $a=-10 x$ satisfies equation (1), hence the particle represent S.H.M.
(d) $a=1 \omega x^3$ also does not satisfy equation (1). So, this does not represent S.H.M.

View full question & answer→

Question 23 Marks

As shown in the picture, A the area of cross-section of the neck of an air chamber of volume is $V$, ball of mass can move up and down in this neck without any friction. Show that when the ball is slightly pressed down and released, it performs simple harmonic motion. Consider the pressure volume variation as isothermal and find the expression for the time period of oscillations.

Answer

Let the volume of a chamber filled with a long neck of internal cross section $= V$.
Let $m= C$ mass of the bullet filled in the neck at position C.
Let $Pa =$ The pressure on both sides of the bullet $=$ Atmospheric pressure
$CD =y$ on increasing the pressure on the bullet, it comes down to the position D .
Where $CD =y$
Thus, the volume of air in the air chamber is less then V and the pressure become more than P .
If $\Delta V =$ reduction in volume in air inside the chamber $= A y$.
Volume distortion $=\frac{\text { Change in volume }}{\text { Volume }}=\frac{\Delta V }{ V }=\frac{ A y}{V}$
Volume modulus of elasticity (E)
$\begin{array}{l}E=\frac{\text { Stress }}{\text { Increase in volume }} \\E=\frac{-P}{\frac{\Delta V}{V}}=-\frac{P}{A y} \times V=-\frac{PV}{A y}\end{array}$
Here the negative sign indicates that an increase in pressure will result in a decrease in the volume of air in the chamber
$P=\frac{-E A}{V} y\ldots\ldots (1)$
Due to excess pressure, the restoring force acting on the bullet is
$F=P \times A=\frac{-EAy}{V} A=\frac{-EA^2}{V} y\ldots\ldots (2)$
Equation (2) with $F =-k y$
$k=\frac{E^2}{V}$
Putting the value in the formula of time period $k$
$\begin{array}{l} T =2 \pi \sqrt{\frac{m}{ k }} \\ T =2 \pi \sqrt{\frac{ mV }{ EA ^2}}\end{array}$

View full question & answer→

Question 33 Marks

If a mass of 0.8 kg is moving in simple harmonic motion starting from equilibrium position. The dimensions of the body of mass 1.0 m and if the period is 11/7 sec then 0.6 m. Find velocity of the particle at displacement and also write the equation of motion.

Answer

Given :
$\begin{array}{l}m=0.8 kg \\A=1.0 m\end{array}$
Period $\quad T=\frac{11}{7} sec \therefore$ Frequency $(n)=\frac{7}{11}$ per sec.
$ \begin{aligned}\text {Displacement}\quad(x) & =0.6 m \\v & =?\end{aligned}$
Equation of the motion $=$ ?
between velocity (v) and displacement ( x ) in simple harmonic motion, their relation :
$\begin{aligned}v & =\omega \sqrt{A^2-x^2}\ldots\ldots (1) \\\omega & =2 \pi n \\& =2 \times \frac{22}{7} \times \frac{7}{11}=4 \text { radians } / sec\end{aligned}$
Putting the value of $v$ in equation (1)
$\begin{aligned}v & =4 \sqrt{(1)^2-(0.6)^2} \\& =4 \sqrt{1-0.36}=4 \sqrt{0.64} \\& =4 \times 0.8=3.2 m / s\end{aligned}$
$\begin{aligned}\text { Displacement } y & =A \sin \omega t \\& =1.0 \sin (4 t) \because \omega=4 \text { radian } / sec \\y & =1.0 \sin (4 t) m\end{aligned}$

View full question & answer→

Question 43 Marks

Show that the particle performs harmonic motion in a parabolic potential well.

Answer

When a body performs simple harmonic motion then it means potential energy at a distance $x$ from the position is P.E. $= U =\frac{1}{2} k x^2$. When we draw graph of potential energy U and displacement X, we get a parabola. This type of potential function is called parabolic potential well because its shape is like a well. Simple harmonic motion is found in this type of potential well. Potential energy $U =\frac{1}{2} k x^2$
Force acting on the particle
$\begin{array}{l}F=-\frac{d}{d x}(U) \\F=-\frac{d}{d x}\left(\frac{1}{2} k x^2\right)=-\frac{1}{2} k \cdot 2 x \\F=-k x\end{array}$
It is clear that the restoring force is proportional to the displacement and its direction is towards the mean potential.
Therefore the body will perform simple harmonic motion.
Acceleration of the particle
$\begin{aligned} a & =-\frac{k}{m} x=-\omega^2 x \\ \text {or}\quad \frac{d^2 x}{d t^2} & =-\omega^2 x\end{aligned}$
and its solution will be $x= A \sin (\omega t+\phi)$. So if any potential function of the object is $U =\frac{1}{2} k x^2$ then the object will perform simple harmonic motion.

View full question & answer→

Question 53 Marks

The time period of a mass hanging from an ideal spring is 2 seconds. If along with it 2 kg if the mass is added and the time period becomes 3 seconds find the value of $m$.

Answer

When a mass of $m$ hung then the time period is T , then
$T=2 \pi \sqrt{\frac{m}{k}}\ldots\ldots (1)$
Where, $k=$ force constant of the spring $(m+2) kg$ 's mass hung then time period is $T ^{\prime}$ so,
$T^{\prime}=2 \pi \sqrt{\frac{m+2}{k}}\ldots\ldots (2)$
Equation (2) divided by (1)
$\frac{T^{\prime}}{T}=\sqrt{\frac{m+2}{m}}$
Given: $T =2$ seconds, $T ^{\prime}=3$ seconds
$\frac{3}{2}=\sqrt{\frac{m+2}{m}}=\frac{9}{4}=\frac{m+2}{m}$
$\text {or}\quad 9 m=4 m+8$
$\Rightarrow5 m=8$
$\Rightarrow m=\frac{8}{5}=1.6 kg$

View full question & answer→

Question 63 Marks

Explain the conservation of mechanical energy in the oscillation of a body suspended from a spring.

Answer

Energy conversion in the motion of a body suspended from a spring further picture is given in which a spring is hanging at the lower end of which a mass is tied, the body is in its equilibrium position. When the body is pulled down a little and released, it starts oscillating up and down, various states of oscillation are shown in the figure.
oscillation of a body suspended from a spring.
When the body is pulled down, the spring stretches. As shown in figure 'A' the work done in stretching the spring started in it in the form of potential energy. Therefore, in the lowest position of the body, the total energy remains present in the form of potential energy in the spring.
When the body is released, the spring starts returning to its normal state and with this the body starts returning towards its equilibrium position. Hence the potential energy of the spring starts converting into kinetic energy of the body. When the body comes to equilibrium, the total energy is in the form of kinetic energy of the body. As shown in figure 'B'.
The body does not remain in the equilibrium position but moves upwards due to inertia due to which the spring starts getting compressed. Again the kinetic energy of the body starts converting into the potential energy of the spring.
In the highest position of the body, its kinetic energy becomes zero and total energy is stored in the form of potential energy of the spring as shown in figure 'C'. In this way, there is mutual transformation of kinetic and potential energies but the value of the total energy remains the same in each case, therefore, mechanical energy is conserved in oscillation.

View full question & answer→

Question 73 Marks

A particle is doing simple harmonic motion. If the velocity of the particle at distances $x_1$ and $x_2$ from the mean position are $v_1$ and $v_2$ respectively, then prove that its time period will be
$ T =2 \pi \sqrt{\frac{x_2^2-x_1^2}{ v _1^2- v _2^2}}. $

Answer

$ v=\omega \sqrt{A^2-y^2} $
$y=x_1, v=v_1$ and $y=x_2, v=v_2$
$ \begin{aligned} v_1 & =\omega \sqrt{A^2-x_1^2} \\ v_1^2 & =\omega^2\left(A^2-x_1^2\right) \\ v_2 & =\omega \sqrt{A^2-x_2^2} \\ v_2^2 & =\omega^2\left(A^2-x_2^2\right) \end{aligned} $
On subtracting equation (2) from (1)
$ \begin{aligned} v_1^2-v_2^2 & =\omega^2\left(x_2^2-x_1^2\right) \\ \omega & =\sqrt{\frac{v_1^2-v_2^2}{x_2^2-x_1^2}} \\ \text { Period } T & =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{x_2^2-x_1^2}{v_1^2-v_2^2}} \end{aligned} $
Hence Proved.

View full question & answer→

Physics 3 Marks Question

Take a timed test